Problem: You have found the following ages (in years) of all 4 lizards at your local zoo: $ 2,\enspace 4,\enspace 2,\enspace 1$ What is the average age of the lizards at your zoo? What is the variance? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 4 lizards at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{2 + 4 + 2 + 1}{{4}} = {2.3\text{ years old}} $ Find the squared deviations from the mean for each lizard. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $2$ years $-0.3$ years $0.09$ years $^2$ $4$ years $1.7$ years $2.89$ years $^2$ $2$ years $-0.3$ years $0.09$ years $^2$ $1$ year $-1.3$ years $1.69$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.09} + {2.89} + {0.09} + {1.69}} {{4}} $ $ {\sigma^2} = \dfrac{{4.76}}{{4}} = {1.19\text{ years}^2} $ The average lizard at the zoo is 2.3 years old. The population variance is 1.19 years $^2$.